Problem 23 - Precalculus Mathematics for Calculus

Question

Sketch the graph of the polynomial function. Make sure your graph shows all intercepts and exhibits the proper end behavior. (GRAPH CANT COPY) $$P(x)=\frac{1}{12}(x+2)^{2}(x-3)^{2}$$

Step-by-Step Solution

Answer

The graph touches at x = -2 and x = 3 and rises to infinity on both ends.

Step 1: Identify the Intercepts

To find the x-intercepts, set the polynomial equal to zero. Given the polynomial \( P(x) = \frac{1}{12}(x+2)^2(x-3)^2 \), we solve \((x+2)^2(x-3)^2 = 0\). The solutions are \( x = -2 \) and \( x = 3 \). These are the x-intercepts. To find the y-intercept, evaluate \( P(x) \) at \( x = 0 \). Thus, \( P(0) = \frac{1}{12}(0+2)^2(0-3)^2 = 4 \times 9 = \frac{1}{12} \times 36 = 3 \). The y-intercept is at \( (0, 3) \).

Step 2: Determine the Multiplicity of Roots

The function has a multiplicity of 2 at both \( x = -2 \) and \( x = 3 \). This indicates that the graph touches the x-axis at these points and does not cross it. The graph also appears parabolic near these intercepts due to the even multiplicity.

Step 3: Assess the End Behavior

To understand the end behavior of the polynomial, notice that the leading coefficient and the degree of the polynomial define it. The polynomial \( \frac{1}{12}(x+2)^2(x-3)^2 \) expands to a degree 4 polynomial \( \frac{1}{12}x^4\) with a positive leading coefficient in expanded form. This means as \( x \to \pm \infty \), \( P(x) \to \infty \). Hence, the graph rises on both ends.

Step 4: Sketch the Graph

Using the information from the previous steps, plot the graph. Start by marking the intercepts at \( (-2, 0) \), \( (3, 0) \), and \( (0, 3) \). The graph touches the x-axis at \( x = -2 \) and \( x = 3 \) without crossing due to the even multiplicity. Additionally, plot the general shape of the function demonstrating that it rises to infinity at both ends because it is a degree 4 function with a positive leading coefficient.

Key Concepts

Intercepts

In polynomial graphs, intercepts are where the graph meets the axes. The x-intercepts occur where the graph crosses or touches the x-axis, meaning the y-value is zero. For a polynomial function like \( P(x) = \frac{1}{12}(x+2)^2(x-3)^2 \), you find the x-intercepts by solving \((x+2)^2(x-3)^2 = 0\).
These solutions are \( x = -2 \) and \( x = 3 \), indicating points where the graph touches the x-axis.

• X-intercepts: \( (-2, 0) \) and \( (3, 0) \)
• The graph will touch the x-axis at these points without crossing it.

To find the y-intercept, set \( x = 0 \) in the polynomial. For our example:\[ P(0) = \frac{1}{12}(0+2)^2(0-3)^2 = 3 \]This gives the point \( (0, 3) \) as the y-intercept, which is where the graph meets the y-axis.

Multiplicity of Roots

Multiplicity indicates how the polynomial is factored and how the graph behaves at its intercepts. It refers to the number of times a particular root appears in the polynomial.
For \( P(x) = \frac{1}{12}(x+2)^2(x-3)^2 \), both \( x = -2 \) and \( x = 3 \) have a multiplicity of 2.

• Even Multiplicity: When the multiplicity of a root is even, like in this case (2), the graph touches the x-axis and "bounces" back rather than crossing it.
• This means the polynomial graph will appear parabolic at these points because it merely touches the axis instead of intersecting it.

End Behavior

The end behavior of a polynomial explains what happens to the graph as \( x \rightarrow \infty \) or \( x \rightarrow -\infty \). Typically, the degree and the leading coefficient provide insights into this behavior.
For the function \( P(x) = \frac{1}{12}(x+2)^2(x-3)^2 \):

• Since it is a degree 4 polynomial, it will rise or fall on either end like a parabola.
• Here, the leading coefficient is positive, indicating that as \( x \rightarrow \pm \infty \), \( P(x) \rightarrow \infty \).
• This means the graph will rise indefinitely on both the left and right ends.

Degree of Polynomial

The degree of a polynomial is the highest power of the variable \( x \) within the polynomial expression. This determines the overall shape and number of x-intercepts.
For the polynomial \( P(x) = \frac{1}{12}x^4 \) after expansion:

• The degree is 4, indicating a quartic polynomial.
• This degree suggests that the graph has a maximum of four x-intercepts, though multiplicity can reduce the number of distinct intercepts observed.
• Additionally, being an even degree polynomial, it shares characteristics of parabolas, given its symmetrical nature around the y-axis as determined by the leading term.

Leading Coefficient

The leading coefficient is the coefficient of the term with the highest degree in a polynomial, and it heavily influences both the graph's shape and direction.
In the polynomial \( P(x) = \frac{1}{12}x^4 \), the leading coefficient is \( \frac{1}{12} \):

• A positive leading coefficient in this even-degree polynomial means that both ends of the graph will point upwards.
• The size of the coefficient (although a fraction here) affects the "stretch" or "compression" of the graph; smaller coefficients result in wider opening curves.
• This ensures that as \( x \rightarrow \pm \infty \), the function continuously rises, confirming the observed end behavior.